Flexural Reinforcing

For bottom bars in major direction,

M_u = 1080.2 k-ft

ϕM_n = ϕRbd² ³ M_u

R ≥ M_u /ϕbd² = (1080.2 k-ft ∙ 12 in/ft)/[0.9 ∙ (180 in) ∙ (26.63 in)²] = 112.8 psi

ρ ≥ 0.001925

A_s ≥ 0.001925 · 180 in · 26.63 in = 9.23 in² ∴ (21) #6 bottom required

With regard to the minimum steel requirement, as we will see below, (11) #4 top will be provided in major direction, thus,

(21 ∙ 0.44 in² + 11 ∙ 0.20 in²) = 11.44 in² > A_s,min

For bottom bars in minor direction,

M_u = (3.85 ksf / 2) ∙ 12.57 ft ∙ 6.75 ft ∙ (6.75 ft / 2) = 551.2 k-ft

By inspection minimum reinforcing will control. Provide bottom bars such that minimum reinforcing limit is met, including top bars.

From calculation below, (11) #4 top will be provided,

A_s > 9.72 in² - (11 · 0.20 in²) = 7.52 in²

Required number of #6 bars = 7.52 in² / 0.44 in² = 18

(18) #6 bottom required

For top bars,

Uplift occurs for Concrete Load Combination thus top bars are required. However, by inspection negative moment induced in footing is small and thus only minimum top bar requirements need be satisfied.

Use (11) #4 top each way to satisfy spacing requirements

Final Design

• 15 '-0 " x 15 '-0 " x 2 '-6 " dp
• (21) #6 long maj & (18) #6 long min, bottom
• (11) #4 long each way, top